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svp [43]
3 years ago
10

(50 points) At a certain high school, 15 students play stringed instruments and 20 students play brass instruments. Five student

s play neither stringed nor brass instruments. What is the probability that a randomly selected student plays both a stringed and a brass instrument?
Mathematics
1 answer:
MakcuM [25]3 years ago
4 0

Answer:

Step-by-stepP(A\cup B)= 0.875

From the diagram, none of the students play both instruments.

This means the two events are mutually exclusive.

P(A\cup B)=P(A)+P(B)

The number of people playing instruments is 20.

P(A\cup B)= \frac{20}{40}  +  \frac{15}{40}

P(A\cup B)= \frac{35}{40}

P(A\cup B)= 0.875

UHHH HI BRAINLIESY HACKS LOL :))((()()()(()(()()()()

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Answer:

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<u>Given line:</u>

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2 years ago
Simplified fraction of 49 out of 112
SVETLANKA909090 [29]
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Vadim26 [7]

Answer:

In this scenario, we will use the <u>femur or the thigh</u> bone length as the explanatory variable.

Step-by-step explanation:

Dependent variables are those variables that are under study, i.e. they are being observed for any changes when the other variables in the model are changed.

The dependent variables are also known as response variables.

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The independent variables are also known as the explanatory variables.

Scientifically it is believed that the length of arm and legs are related and basically grow at the same time.

So, in this case the explanatory variable can either of the two bone lengths.

Thus, the complete statement is:

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4 years ago
Reggie Means invested part of $31,000 in municipal bonds that earn 7.5% annual simple interest and the remainder of the money in
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Answer:

Principal Invested in Municipal Bonds is $16,000

Principal Invested in Corporate Bonds is $15,000

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If;

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From equation 1 we have;

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By putting the values of a in Equation 2 we get;

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